posted by Cindy on .
A man drags a table 4.25 m across the floor, exerting a constant force of 54.0 N, directed 28.0° above the horizontal.
I have found the work done by the applied force to be 202.6 J but I can't find the second part of the question which is below.
(b) How much work is done by friction? Assume the table's velocity is constant.
(b) Multiply the friction force by the distance the table moves. It will be a negative number.
Since the table does not accelerate, you know that the friction force must balance the horizontal component of the applied force. The answer to (b) is MINUS the value of the answer to (a)